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Any idea what to do for 3b and 4?

3 Answers

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For 3a, find the midpoint of PQ. Then, find the line passing through PQ with a slope equal to the negative reciprocal of the slope of PQ.

For 3b, find the equation of the line, then solve with 3a.

Step-by-step:

The midpoint of PQ = (6, 2), and the slope is -4/3.

3a passes through (6, 2) and has a slope of 3/4.

y = (3/4)x - 5/2

The line passing through (20, 2) with a slope of -1/8 is

y = (-1/8)x + 9/2

The intersection of these lines:

(3/4)x - 5/2 = (-1/8)x + 9/2

(7/8)x = 7

x = 8

y = (-1/8)(8) + 9/2

y = 7/2

(8, 7/2)
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For 3b, I tried using y-b=m(x-a) with coordinates (20,2) and gradient -1/8. I ended up getting x+8y=0 but do not know if I can pull coordinates for S this way.
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for 4, rearrange so that sinx = -3/4. Then, solve for x in the domain x e [0, 360]

x1 = 360 - |arcsin(-3/4)|

x2 = 180 + |arcsin(-3/4)|

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